I Hysteresis of a Compressed Solid

[Edsel]

I’m not really sure where to ask this, but here goes:

I am trying to find a topic to read on which explains the behavior of brass during reloading (an activity that’s part of recreational shooting).

Shooters who handload recycle brass by refilling brass cases with new primers, smokeless powder, and bullets. But said brass cases need to be sized down to fit with sizing dies prior.

Let’s say that brass case fired from Person A’s hunting rifle gets ballooned by +0.002” longer than the chamber. The sizing die, set to a fixed volume suiting the tight chamber of Person A’s rifle, compresses the brass case down to -0.001” shorter than the chamber.

Now, brass fired from Person B’s loose chambered hunting rifle gets ballooned by +0.006” longer than the chamber of Person A’s rifle. Using Person A’s sizing die set to the same prior fixed volume suited to his rifle’s tight chamber, one can only compress Person B’s brass case down to +0.002” longer than the tighter chamber of Person A’s rifle. This is achieved after multiple compressions, gradually going from +0.004”, then +0.003”, and finally +0.002”… But there appears to be an asymptotic limit of sorts at +0.002”, never truly achieving -0.001”. Person B’s brass cases cannot be sized to fit Person A’s rife, unless Person A reduces the volume of his sizing die.

A crude analogy would be as follows: A 200 lb individual can only get down to 70 lbs after radical surgery, while a 120 lb individual can get down to 50 lbs after identical radical surgery. The “limits” for each, if you will - are different.

I imagine that this could be explained by some Hysteresis Curve of sorts, but am at a loss at finding the specific topic which explains this behavior.

Is there any basic property with accompanying equations which readily explains such?

Thanks in advance.

 

[Arjan82]

I'm not acquainted at all with what you are describing here. Do you have some photos or figures which explain the ballooning, and the die process? Why won't it go to -0.001”? I mean, what happens if you try to do that? You use the same die and the same process but it just won't? Or will it deform the brass case? Or something else?

The only thing I can imagine now is that the thickness of the brass case changes slightly for person B because of the larger ballooning and the die process is not able to undo that without buckling/warping the brass case, or something like that.

 

[Baluncore]

I imagine that this could be explained by some Hysteresis Curve of sorts, but am at a loss at finding the specific topic which explains this behavior.

Is there any basic property with accompanying equations which readily explains such?

https://en.wikipedia.org/wiki/Stress–strain_curve

The metal requires a force to move it through elastic deformation, storing energy in the structure, before it reaches the yield point, from which point on, it yields by plastic deformation. When the force is removed, the metal recovers elastically, releasing the elastic energy, but to a different size because the plastic deformation remains.

The term coining is applied to the process where significantly higher forces change the internal structure of the metal. That effectively heats and anneals the material, releasing the energy stored in the elastic deflection, so there is then no elastic spring back.

After firing, the brass cartridge will have been explosively coined to fit the chamber. To reuse a cartridge, it will need to be drawn, to reduce the diameter to meet the required specification. The drawing process first takes the diameter to a lower value than the specification, knowing that the material will elastically spring back by a small percentage, to the required size. The length will then be trimmed with a cutter.

Each time a cartridge is resized, the wall of the cartridge will be thinned, with the excess material being lost during the trim to length. The cartridge must be discarded once it cannot be drawn, or its weight falls below a safe limit.

 

[Edsel]

I'm not acquainted at all with what you are describing here. Do you have some photos or figures which explain the ballooning, and the die process? Why won't it go to -0.001”? I mean, what happens if you try to do that? You use the same die and the same process but it just won't? Or will it deform the brass case? Or something else?

The only thing I can imagine now is that the thickness of the brass case changes slightly for person B because of the larger ballooning and the die process is not able to undo that without buckling/warping the brass case, or something like that.

Here’s a primer.

 

[Edsel]

Apparently, my search terms weren’t optimal.

Instead of “hysteresis of solids,” I should have searched for “forming and springback.”

 

[Edsel]

https://ahssinsights.org/forming/springback/

This laid out the framework…

http://li.mit.edu/Stuff/RHW/Upload/49.pdf

…while this gave me what I really needed.

Hysteresis does occur - and Vf is influenced by Vi.

More “springback” occurs when your initial volume is larger.

Hence the inability to adequately size the more bloated brass using the same fixed compression volume.

The only way to do so would be to adjust the die to reduce the compression volume further.